Journal of Algebra 347 (2011) 133–142

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Journal of Algebra

www.elsevier.com/locate/jalgebra

Groups that involve finitely many primes and have allsubgroups subnormal

Howard Smith

Department of Mathematics, Bucknell University, Lewisburg, PA 17837, USA

a r t i c l e i n f o a b s t r a c t

Article history:Received 28 April 2011Available online 4 October 2011Communicated by E.I. Khukhro

Keywords:Subnormal subgroups

It is shown that if G is a group with all subgroups subnormal, andif the torsion subgroup of G involves just finitely many primes,then G has a normal nilpotent subgroup N such that G/N isa Chernikov group.

© 2011 Elsevier Inc. All rights reserved.

1. Introduction

Let N1 denote the class of groups with all subgroups subnormal. It was proved in [4] that a peri-odic group G in N1 is nilpotent-by-divisible Chernikov, that is, G has a normal nilpotent subgroup Nsuch that G/N is a divisible Chernikov group. Most of the proof is concerned with showing that theresult holds when G is a p-group for some prime p; the general case is then an easy consequence ofRoseblade’s Theorem [16], which states that a group G in which every (finitely generated) subgroupis subnormal of bounded subnormal defect r is nilpotent of class bounded in terms of r only. In eachof the papers [3] and [19] it is shown that a torsion-free group in N1 is nilpotent. (The proof in [3] isconsiderably shorter than that in [19].) Certainly there are non-nilpotent groups G in the class N1. Thebest-known such examples are those constructed by Heineken and Mohamed [10] – these centrelessmetabelian p-groups have all proper subgroups both subnormal and nilpotent. The reader is referredto Chapter 6 of [12] for further discussion of the Heineken–Mohamed groups; in addition, furthersuch examples, of unbounded derived length, are constructed in [13]. Later it was shown by Möhres[15] that every N1-group is soluble; this article was one of a sequence of six important papers byMöhres on N1-groups, and the interested reader is invited to consult [15].

Now, there is no possibility of showing that every N1-group is nilpotent-by-Chernikov, as maybe seen by considering the two (infinite) families of examples constructed in [17]. These groupsare hypercentral of length exactly ω + 1, and although they are all periodic-by-(locally cyclic) (and

E-mail address: howsmith@bucknell.edu.

0021-8693/$ – see front matter © 2011 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2011.09.014

134 H. Smith / Journal of Algebra 347 (2011) 133–142

every group in one of these two families even has finite Prüfer rank), none of them is nilpotent-by-Chernikov – we recall that a group G is said to have Prüfer rank at most r if every finitely generatedsubgroup of G is r-generated. More examples of N1-groups that are hypercentral of length exactlyω + 1 are provided in [21], where each group has infinite torsion-free rank but every subgroup thathas finite torsion-free rank is nilpotent. The only known non-nilpotent N1-groups of which I amaware, beyond those referred to (at least implicitly) above and besides those that can be constructedin a trivial manner from these (e.g. by taking certain direct products) are the derived-length 3 ex-amples constructed in [2] and some more p-groups constructed in the doctoral thesis of Möhres –these latter groups are based on the Heineken–Mohamed examples and are (elementary abelian)-by-divisible Chernikov.

In view of the results and examples described above, it is reasonable to ask whether everyN1-group G is nilpotent-by-(finite rank). This question is not answered in the present article, butthe special case where the torsion subgroup of G is a p-group is settled in the affirmative. Notice thatthis in itself provides a generalization of what is really the main result of Casolo in [4], namely thatconcerned with p-groups – we have already observed that passing from p-groups to arbitrary peri-odic groups requires no more than an easy application of Roseblade’s Theorem. In fact, we establishthe following.

Theorem. Let G be a group with all subgroups subnormal, and suppose that the torsion subgroup of G is aπ -group, where π is some finite set of primes. Then G is nilpotent-by-divisible Chernikov.

In order to prove this theorem we require a considerable number of results from the literature onN1-groups. In addition, we require some auxiliary results that will allow us to proceed with the mainpart of the proof, the reduction to a very special case, carried out in Section 3. These auxiliary resultsare presented in Section 2, along with some further results that are introduced in order to reduce firstto the case where our N1-group G is metabelian, has torsion subgroup a p-group and enjoys certainother useful properties.

Finally in this introduction we observe that, in the context of our discussion, nilpotent-by-(finiterank) and nilpotent-by-divisible Chernikov are in fact identical properties. To see this, suppose that Gsatisfies the hypotheses of the theorem and that G has a normal nilpotent subgroup N with G/Nof finite rank. Since G/N is locally nilpotent there is a finitely generated subgroup F of G such thatG is the isolator of N F , that is, every element of G has a non-zero power in N F . (See [9] for thebasic properties of isolators in locally nilpotent groups.) Since N is nilpotent and normal and F isnilpotent and subnormal, N F is also nilpotent, as therefore is its normal closure M , say, in G . (Thislatter result follows immediately from the main result of [5], though in this special case it is farmore elementary.) G/M is the direct product of its π -component G1/M , which has finite rank andis therefore Chernikov, and its π ′-component G2/M , and since G2 is π ′-free and M is nilpotent, wehave G2 nilpotent and hence G nilpotent-by-(Chernikov-π ). Now a Chernikov group that is also aBaer group is center-by-finite and therefore has finite derived group, and since a nilpotent-by-finiteBaer group is nilpotent we have G nilpotent-by-divisible Chernikov, as required.

2. Some preliminaries

We begin with an easy result that will allow us to factor by certain normal subgroups during thecourse of the proof of the theorem.

Lemma 1. Let G be a locally nilpotent group, T the torsion subgroup of G, and suppose that T is a π -groupand G/T is nilpotent. Let N be a normal nilpotent subgroup of G and let U/N ′ be the π ′-component of thetorsion subgroup of G/N ′ . If G/U is nilpotent-by-(finite rank) then so is G.

Proof. With the hypotheses stated, there is a normal subgroup K of G with G/K of finite rank, suchthat U � K and K/U is nilpotent. Thus there is a positive integer c such that γc(K ) � T ∩ U ; since Tis a π -group and U is π ′ mod N ′ we see that γc(K ) � N ′ , that is, K/N ′ is nilpotent. By a well-knownresult of P. Hall [8] we have K nilpotent, and the result follows. �

H. Smith / Journal of Algebra 347 (2011) 133–142 135

For our next result we need some notation from [4]. Let p be a prime and let P be an abelianp-group. P is said to be small if it is the direct product of a divisible Chernikov subgroup and asubgroup of finite exponent; otherwise P is large. (Thus (with the usual notation) P large means thatΩi+1(P )/Ωi(P ) is infinite for all i � 0.)

The following result establishes a special case of our theorem; its proof follows almost exactly thatof Theorem 1 of [4] (which was concerned entirely with p-groups).

Lemma 2. Let p be a prime, G a group with all subgroups subnormal, and suppose that H, K are subgroupsof G such that K � H � G, K is nilpotent and G/H and H/K are p-groups, with G/H abelian and H/Kdivisible Chernikov. Suppose further that every subgroup S of G that has a normal nilpotent subgroup R withS/R abelian is nilpotent-by-Chernikov. Then G is nilpotent-by-Chernikov.

Proof. Suppose the result false and let G be a counterexample with the rank r of H/K minimal forsome H and K . By hypothesis K < H and so r > 0. Since K G is nilpotent (e.g. by [5]) we may assumeK � G . Now H/K is central in the Baer group G/K and we may apply Lemma 10 of [4] to the p-groupG/K . If G/H is small then there is a normal subgroup N/K of G/K that has finite exponent and issuch that G/N is divisible Chernikov. By the main result of [18] N is nilpotent, and since G/N isabelian we have the contradiction that G is nilpotent-by-Chernikov. Thus G/H is large.

Let W /K be a normal subgroup of G/K that is maximal such that W ∩ H = K . Then W ′ � Kand so W is nilpotent-by-abelian and there is a normal subgroup Y of W such that K � Y andW /Y is Chernikov. We may choose Y so that W /Y is divisible – observe too that Y is G-invariantsince [W , G] � K . If G/H W is small then G/Y is (divisible-Chernikov)-by-(finite exponent), and byLemma 10 of [4] we see that G/Y is (finite-exponent)-by-divisible Chernikov. Again by [18] we obtainthe contradiction that G is nilpotent-by-Chernikov. Thus G/H W is large, and by Lemma 10 of [4]there is a large abelian subgroup X/W of G/W such that X ∩ H W = W . Thus X ∩ H = K and X/Kis abelian, and again by hypothesis there is a normal nilpotent subgroup U of X such that K � U andX/U is divisible Chernikov. Note that U G is nilpotent and contained in HU . Since K � U , H/H ∩ U G isdivisible and hence of rank r, by minimality. Thus (H ∩ U G)/K is finite. But U G/U = U (H ∩ U G)/U ∼=(H ∩ U G)/(H ∩ U ) = (H ∩ U G)/K , so U G/U is finite. Since XU G/U G is divisible it is central in (thenilpotent p-group) G/U G ; hence X G = XU G .

Next we observe that X G/X ∼= U G/(X ∩ U G) = U G/U , which is finite. Thus there is a positiveinteger n such that M := (X G)pn

is contained in X , and so X contains the normal subgroup W Mof G . Then W M ∩ H = K , and by choice of W we have M � W and hence X pn � W , contradictingthe fact that X/W is large. This completes the proof of the lemma. �

We now have what we need in order to effect the partial reduction referred to above, which isgiven as our proposition below, while the lemmas that follow will be required later on in the proofof the theorem. The proof of the first of these is very similar to that of Proposition 1.K .2 of [11], andhere we omit a few of the details and refer the interested reader to [11].

Lemma 3. Let G be a group, A a normal abelian subgroup of G such that G/A is locally finite, and let c, r bepositive integers. Suppose that, for every finitely generated subgroup F of G, there is a normal subgroup N F ofA F such that A � N F , N F is nilpotent of class at most c, and A F/N F has rank at most r. Then there is a normalsubgroup N of G such that A � N, N is nil-c and G/N has rank at most r.

Proof. Let Σ be the local system of G consisting of all subgroups A F of G , with F finitely generated.For each L in Σ , let I L be the set of all normal nil-c subgroups X of L such that A � X and L/X hasrank at most r. If M ∈ Σ and L � M then, given X ∈ IM , we see that X ∩ L ∈ I L , so we have a map(M → L) from IM to I L , and each I L is non-empty by hypothesis. Thus we have an inverse system{I L, (M → L); L ⊆ M, M ∈ Σ} of finite non-empty sets (since A F/A is finite for every finitely generatedsubgroup F of G), and so its inverse limit Λ is also non-empty. Choose (XL) in Λ, where XL ∈ I L foreach L ∈ Σ , and let X be the union of the XL over all L in Σ . The subgroups XL form a local systemfor the normal nil-c subgroup X of G , and of course A � XL � X for each L. If F is an arbitrary finitely

136 H. Smith / Journal of Algebra 347 (2011) 133–142

generated subgroup of G then L := A F ∈ Σ and F XL/XL has rank at most r, so F X/X and hence G/Xhas rank at most r. �Lemma 4. Let G be a group with all subgroups subnormal, and suppose that the torsion subgroup T of G isa p-group for some prime p and that G/T is abelian. Suppose too that G has a normal abelian subgroup Acontaining T such that G/A is periodic.

(a) For a given positive integer k, if G is not nilpotent-by-(finite rank) then neither is G/Ωk(T ).

(b) If V denotes the intersection of the subgroups T pk, k = 1,2, . . . then V is central in G (hence if G/V is

nilpotent-by-(finite rank) then so is G).

Proof. For part (a), let S = Ωk(T ) and suppose for a contradiction that there is a normal subgroup Hof G with S � H, H/S nilpotent and G/H of finite rank. Then AH/S is also nilpotent, so there is apositive integer c such that [A, c H] � S , and then [A pk

, c H] = [A, c H]pk � S pk = 1. Thus A pk ∩ H �Zc(H) and H/Zc(H) is periodic and hence nilpotent-by-Chernikov, by [4]. But then G is nilpotent-by-Chernikov, a contradiction. For part (b), let g be an arbitrary element of G and observe that thereare positive integers m and c such that g pm

centralizes T and T 〈g〉 is nilpotent of class c. Just as inthe proof of Lemma 2.2 of [20] (for example) it is easily shown that [T p(c−1)m

, g] = 1, and the resultfollows. �Lemma 5. Let K = T 〈g〉 be a nilpotent group of class c, where T is a normal abelian subgroup of K of expo-nent pl . Then [T , g pm ] = 1, for some m depending only on c and l.

Proof. If t ∈ T then |t| � pl and 〈t〉〈g〉 is generated by the elements [t, i g], i = 0,1, . . . , c − 1 andhence has order at most pcl . The result follows easily. �Lemma 6. Let p be a prime, G a nilpotent group with torsion subgroup T of finite exponent pl and central in G

and G/T abelian. Then G pl+1 ∩ T = 1.

Proof. This is just a special case of Lemma 7 of [5]: for x, y ∈ G we have (xy)pl+1 = xpl+1ypl+1 ×

[x, y](pl+1

2 ) = xpl+1ypl+1

(since pl divides(pl+1

2

)). Thus if g ∈ G pl+1 ∩ T then g = hpl+1

for some h whichmust also lie in T , hence g = 1, as required. �Proposition. Let G∗ be a group with all subgroups subnormal and suppose that the torsion subgroup T ∗ of G∗is a π -group for some finite set π of primes. Suppose too that G∗ is not nilpotent-by-(finite rank). Then G∗ hasa section G with the following properties.

G is not nilpotent-by-(finite rank), the torsion subgroup T of G is an abelian p-group for some p ∈ π andG/T is abelian, and there is an abelian subgroup A of G that contains T and is such that A/T is free abelianand G/A is a p-group.

Proof. For ease of notation we shall refer to our group as G at the outset, and re-label appropriatelyas we go along. Firstly, we know from [6] that the class of groups that are nilpotent-by-(finite rank)is countably recognizable, and so we may assume that G is countable. Since G is locally nilpotent, ittherefore has a torsion-free subgroup H whose isolator in G is G itself (see, for example, Proposition 3of [7]). By [3] (or [19]) H is nilpotent, and so H G is nilpotent, by the main result of [5]. By [3] wehave G/T nilpotent, and from [4] we know that T has a normal nilpotent subgroup S such that T /S isdivisible Chernikov; since SG is also nilpotent we may choose S to be G-invariant. Then A := S H G is anormal nilpotent subgroup of G , and an application of Lemma 1 allows us to factor by an appropriatesubgroup containing A′ and retain our original hypotheses – here we may have lost the torsion-freeness of H , but that is no longer required. A is now a normal abelian subgroup of G with G/Aperiodic, and since G/T is torsion-free and abelian-by-periodic it is abelian; also T A/A is divisibleChernikov since S � A.

H. Smith / Journal of Algebra 347 (2011) 133–142 137

Next, for each prime p in π , let V p denote the p′-radical of T , and suppose that for each such pthere is a normal subgroup N p/V p of G/V p such that G/N p has finite rank. If N denotes the inter-section of all the N p then G/N has finite rank and N is nilpotent. By this contradiction there is somep ∈ π such that G/V p is a counterexample to the theorem, and we may factor and hence assumethat T is a p-group. Now G/A is periodic and is the direct product of its p-component G1/A andits p′-component G2/A, and since T is a p-group it follows that G2 is abelian, so we may replace Gby G1 if necessary and assume that G/A is a p-group.

Since T A/A is divisible Chernikov we may apply Lemma 2 to deduce that G has a non-(nilpotent-by-Chernikov) subgroup G3 that is nilpotent-by-abelian; since G3 inherits the relevant conditionson G we may assume that G3 = G , which therefore has a normal nilpotent subgroup M with G/Mabelian; since AM is nilpotent we may suppose that M contains A. As T M/M is divisible we haveG/M = G4/M × T M/M for some subgroup G4 that is also a counterexample because T M/M isChernikov. We may replace G by G4 and hence assume that T is contained in M . By a special case ofLemma 1, G/M ′ is again not nilpotent-by-Chernikov, and since M ′ � T we retain all of our hypothesesif we factor by M ′ i.e. assume that M is abelian. Finally, let B/T be a free abelian subgroup of M/Twith G/B periodic. Writing G/B as the direct product of its p- and p′-components, we may pass onceagain to an appropriate subgroup of G to assume that G/B is a p-group, and this concludes the proofof the proposition. �3. Proof of the theorem: reduction to a special case

Let us denote by X the class of groups that are nilpotent-by-(finite rank), and recall that everyX-group G with all subgroups subnormal and π(T (G)) finite is nilpotent-by-(divisible Chernikov).Throughout this section, G is an N1-group satisfying the hypotheses of the theorem, and we aresupposing for a contradiction that G is not in X. We shall also suppose that G has the structuredescribed in the proposition. One further property that we shall assume at the outset is the following.

There is a positive integer d and a finitely generated subgroup F of G such that every non-X-subgroup Sof G that contains F is subnormal of defect at most d in G.

This assumption is justified by an application of Lemma 2 of [14], a result that is (essentially)due to Brookes [1]. Since A F is nilpotent and normal in G we may assume, by (a special case of)Lemma 1 that A F is abelian; since A F/T is free abelian, we may also assume that F � A. There is asubgroup F ∗ of A with F of finite index in F ∗ and T F ∗/T a direct factor of A/T . Replacing F by F ∗if necessary we may write A/T = T F/T × A1/T for some subgroup A1 of A. Also, since the torsionsubgroup of F is finite and hence contained in Ωk(T ) for some integer k, Lemma 4 allows us to factoronce more and hence assume that F is torsion-free.

Let c1 < c2 < · · · , r1 < r2 < · · · be sequences of positive integers. Our goal in this section is toconstruct a non-X section H of G such that the torsion subgroup T ∗ of H is the direct productof finite H-invariant abelian subgroups Ti , i = 1,2, . . . and H/T ∗ is the direct product of abeliansubgroups Fi T ∗/T ∗ , i = 0,1,2, . . . (where F0 = F ) such that each Fi is finitely generated and, inparticular and for each i � 1, Ti Fi is not (nilpotent of class ci)-by-(rank ri). We shall see that F1 isstraightforward to construct and we shall describe the construction of F2 in considerable detail andthat of F3 in a little less detail; this should suffice to indicate how the remaining subgroups Fi are tobe constructed.

We begin by letting J1 = G1 = IG(A1), the isolator in G of A1. (At subsequent steps in our re-duction the subgroups designated J i and Gi will differ from each other.) Then G/G1 has finite rankand so G1 /∈ X. As G1/A1 is locally finite, we may apply Lemma 3 to deduce that there is a finitelygenerated subgroup F1 of G1 such that, for every (normal) nil-(c1 + 2) subgroup M of A1 F1 thatcontains A1, A1 F1/M does not have rank at most r1; we may choose F1 so that it is generated byfree generators mod T . Let D0 = ⋂∞

n=1 T pn, the finite residual of T . By Lemma 4, D0 is central in G1.

There are just finitely many subgroups M of A1 F1 such that A1 � M; if M is any such subgroup andA1 F1/M has rank at most r1 then 1 �= γ(c1+3)(M) � [T , c1 M, G1] and so [T , c1 M] � D0. Since the nor-

138 H. Smith / Journal of Algebra 347 (2011) 133–142

mal closure in A〈F , F1〉 of every element of T is finite, we see that there is a finite 〈F , F1〉-invariantsubgroup U1 of T such that [U1, c1 M] � D0 for any such subgroup M of A1 F1.

Now T1 := 〈F , F1, U1〉 ∩ T is finite and hence contained in Ωk1 (T ) for some integer k1, and wesee that G1/D0Ωk1 (T ) /∈ X. Furthermore, there is a positive integer l1, and we choose l1 > 1, such

that T1 ∩ T pl1 � D0; since every finitely generated subgroup of G is finite modulo A we may choosean 〈F , F1〉-invariant subgroup S1 of finite index in T and containing T pl1 (and hence D0) such thatT1 ∩ S1 � D0.

There is a subgroup A2/T of A1/T such that (F1 ∩ A)T /T × A2/T has finite index in A1/T . Let J2 =IG(A2); then G/ J2 has finite rank and J2 /∈ X. However, J2/T pl1 = J2/(A pl1 ∩ T ) = J2/(( J2 ∩ A pl1

)∩ T )

and, since J2/T is abelian and J2/( J2 ∩ A pl1) is periodic and hence in X (by [4]), it follows that

J2/T pl1 ∈ X, so there is a nilpotent subgroup K2/T pl1 of J2/T pl1 with T � K2 and J2/K2 Chernikov.

By Lemma 5 there is a positive integer m1, which we choose so that m1 � l1, such that K pm1

2 cen-

tralizes T /T pl1 (and hence normalizes S1). By Lemma 6 (applied to the group T K pm1

2 /T pl1 ), we have

K p2m1

2 ∩ T � T pl1(� S1). Let G2 = K p2m1

2 S1 and note that S1 is the torsion subgroup of G2.If G2 ∈ X then K2 is nilpotent-by-Chernikov-by-(finite exponent) and so J2 (which is nilpotent-

by-abelian since T � J2) is nilpotent-by-small. But a nilpotent-by-(finite exponent) N1-group is nilpo-tent [18], and we have the contradiction that J2 ∈ X. It follows that G2 /∈ X. Observe that G2 � J2

and G2 ∩ 〈F , F1〉T � G2 ∩ T = S1 � 〈G2, F , F1〉.Recall that T1 � Ωk1 (T ), and set E1 := D0(S1 ∩ Ωk1 (T )). Then T1 ∩ E1 � T1 ∩ S1 � D0. Now let

D1/E1 be the finite residual of T /E1 and hence of S1/E1. If G2/D1 has a normal nilpotent subgroupG∗

2/D1 with G2/G∗2 Chernikov then, by two applications of Lemma 4, we have first that G∗

2/E1 nilpo-tent and then the contradiction that G2 ∈ X. Thus G2/D1 /∈ X. We are in a position at last to continuewith our construction.

By Lemma 3, there is a finitely generated subgroup F2 of G2, which we may choose so that F2 isgenerated by free generators mod T , such that for every subgroup M/D1 of A2 F2/D1 with A2 � Mand A2 F2/M of rank at most r2, M/D1 is not nilpotent of class at most c2 +1. Since F ′

2 � S1 it followsthat [S1, c2 M] � D1 for all such M , and there is a finite 〈F , F1, F2〉-invariant subgroup U2 of S1 suchthat, again for all such M , [U2, c2 M] � D1. Note that S1 is normalized by F2 since it is normalizedby G2.

Now T2 := 〈F , F1, F2, U1, U2〉 ∩ S1 is finite and hence contained in Ωk2 (T ) for some k2, and there

exists l2 > l1 such that T1T2 ∩ T pl2 � D1. Then there is an 〈F , F1, F2〉-invariant subgroup S2 of finiteindex in S1 and containing D1 such that T pl2 � S2 and T1T2 ∩ S2 � D1. Let E2 = D1(S2 ∩Ωk2 (T )) andlet D2/E2 be the finite residual of T /E2 and hence of S2/E2.

There is a subgroup A3/T of A2/T such that (F2 ∩ A)T /T × A3/T has finite index in A2/T ; setJ3 = IG(A3) and note that J3 /∈ X but J3/T pl2 ∈ X. Let K3/T pl2 be a normal nilpotent subgroup of

J3/T pl2 with T � K3 and J3/K3 Chernikov, and choose m2 � l2 so that K pm2

3 centralizes T /T pl2 (and

hence normalizes S2). Then K p2m2

3 ∩ T � T pl2(� S2), and we note that S2 is the torsion subgroup

of G3. Set G3 = K p2m2

3 S2. We have G3 ∩ 〈F , F1, F2〉T � G3 ∩ T = S2 � 〈G3, F , F1, F2〉, and S1 is also

normalized by G3, since [S1, G3] � T pl2 � T pl1 � S1. Furthermore, G3/D2 /∈ X.We may now repeat the above procedure, beginning with a finitely generated subgroup F3 of G3,

generated by free generators mod T , such that for every subgroup M/D2 of A3 F3/D2 with A3 � Mand A3 F3/M of rank at most r3, M/D2 is not nilpotent of class at most c3 + 1. There is a finite〈F , F1, F2, F3〉-invariant subgroup U3 of S2 such that, for every such subgroup M/D2, [U3, c3 M] � D2.Set T3 = 〈F , F1, F2, F3, U1, U2, U3〉 ∩ S2, which is in Ωk3 (T ) for some k3, and choose l3 > l2 such

that T1T2T3 ∩ T pl3 � D2. There is an 〈F , F1, F2, F3〉-invariant subgroup S3 of finite index in S2 andcontaining D2 such that T pl3 � S3 and T1T2T3 ∩ S3 � D2. Note that S2 and S1 are also normalizedby F3.

Continuing in this manner, and with the obvious notation, we obtain the following.

H. Smith / Journal of Algebra 347 (2011) 133–142 139

(a) An ascending sequence of integers 2 � l1 < l2 < · · · and a descending chain T = S0 > S1 > S2 > · · ·of subgroups, each of finite index in T , with T pli � Si for each i � 1.

(b) A sequence of positive integers k1,k2, . . . and ascending sequences of subgroups D0 � D1 � · · · ,E1 � E2 � · · · of T , with Di � Si for all i (and hence Di � S j for all i, j), such that D0 is the finiteresidual of T and, for each i � 1, Di/Ei is the finite residual of T /Ei , with Ei = Di−1(Si ∩Ωki (T )).

(c) A sequence U1, U2, . . . of finite subgroups of T , with Ui � Si−1 for each i, and a sequenceF1, F2, . . . of finitely generated subgroups of G that generate their direct product mod T , witheach Fi generated by free generators mod T , such that, for each i � 1, Ui is 〈F , F1, . . . , Fi〉-invariant and there is no normal subgroup L of Fi with Fi/L of rank at most ri and [Ui, ci L] �Di−1. (This statement requires for its justification the fact that [Ui, ci L] = [Ui, ci AL] for any sub-group L of Fi .)

(d) Ui � Ti � Ωki (T ) for each i � 1, where Ti = 〈F , F1, . . . , Fi, U1, . . . , Ui〉 ∩ Si−1; Ui ∩ Si � Ti ∩ Si �T1T2 . . . Ti ∩ Si � Di−1.

Now let J∗ = 〈F , F1, F2, . . .〉. It is clear that T J∗ is not in X, so we may as well suppose thatG = T J∗ . Each Si is normal in G , for Si is chosen so as to be normalized by F j for each j = 1, . . . , i,while if j > i then we have F j � G j (again with the notation suggested), and hence [Si, F j] �[T , G j] � T pl( j−1) � T pli � Si . Also, whenever i < j we have [Ti, F j] � [Ωki (T ), F j] � Ωki (T )∩[T , F j] �Ωki (T ) ∩ Si � Ei .

Let N = ⋂∞i=1 Si . By (d) and the above we have T1T2 . . . Ti ∩ Si � Di−1 � N for every i and

[Ti, F j] � Ei � N whenever i < j. It follows that, mod N , the Ti are all normal in G and generatetheir direct product T ∗ , say.

Suppose next that G/N has a normal nil-c subgroup B/N with G/B of rank r. Choose any i suchthat ci > c and ri � r, and let C = B ∩ (Ui Fi). Then C Ui is nil-(c + 1) mod N (by Fitting’s Theorem,since C and Ui are normal in Ui Fi ) and so [Ui, c+1C Ui] � N � Si . Hence [Ui, ci C ∩ Fi] � Ui ∩ Si � Di−1

and Fi/(C ∩ Fi) has rank at most ri . This contradiction shows that G/N /∈ X.Next we claim that the following holds.

(e) For all i, j with 1 � i � j, [F , F j] � T j and [Fi, F j] � T j .

Firstly we note that 〈F , F1〉 ∩ T � T1, by definition. With the notation suggested earlier, for each

j > 1 we have that F j is contained in G j = K p2m j−1

j S j−1. Here K j contains T and is therefore normal

in G , and it follows that [〈F , F1, . . . , F j〉, F j] � [〈F , F1, . . . , F j〉, K p2m j−1

j ]S j−1 � (T ∩ K p2m j−1

j )S j−1 �(T pl j−1

)S j−1 � S j−1. (In fact, S j−1 = T ∩ G j , the torsion subgroup of the normal subgroup G j of G ,but the previous calculations may serve as a reminder of some of our notation.) By definition of T j itnow follows that [〈F , F1, . . . , F j〉, F j] � T j , and the claim is established.

Now recall that F and the Fi generate their direct product mod T and that each of these sub-groups is generated by free generators mod T . If x ∈ J∗ then there is an element σ of ( J∗)′ (whichis contained in T ), elements y1, . . . , yk of J∗ that freely generate 〈y1, . . . , yk〉 mod T and integersn1, . . . ,nk such that x = yn1

1 . . . ynkk σ . If also x ∈ T then it follows that yn1

1 . . . ynkk ∈ T and so ni = 0

for all i, which gives x ∈ ( J∗)′ . Hence T ∩ J∗ = ( J∗)′ , and (e) implies that ( J∗)′ is contained in T ∗ .Factoring by N and replacing T by T ∗ if necessary, we may assume the following.

The torsion subgroup T of G is the direct product of the finite G-invariant abelian p-subgroups Ti and G/Tis the direct product of T F/T and the factors T Fi/T , i = 1,2, . . . , with the additional conditions (e) aboveand the property that, for each i � 1, there is no normal subgroup L of Fi with [Ti, ci L] = 1 and Fi/L ofrank at most ri . Furthermore, [Ti, F j] = 1 whenever i < j (since [Ti, F j] � Ei(� N)), and there is a positiveinteger d such that every non-X-subgroup of G that contains F has subnormal defect at most d in G.

This completes the reduction outlined at the beginning of Section 3.

140 H. Smith / Journal of Algebra 347 (2011) 133–142

4. Conclusion of the proof of the theorem

Suppose that G satisfies the hypotheses of the theorem but that G is not in X, and suppose (as wemay) that G satisfies the conditions described at the end of the previous section. As before we areassuming that G has an abelian normal subgroup A containing T F such that A/T is free abelian andFi/(A ∩ Fi) is a finite p-group for every i � 1.

Lemma 7. G is residually finite, and if N is the direct product of G-invariant subgroups Ni , with Ni � Ti foreach i, then G/N is residually finite. Furthermore, every residually finite section of G that is in X is nilpotent.

Proof. If t ∈ G \ T then, since G/T is free abelian, there is a normal subgroup K of finite index in Gsuch that t /∈ K . If 1 �= t ∈ T then t has non-trivial projection onto some T j . Let M = 〈Ti: i �= j〉, L =〈Fi: i > j〉. By (e) above we have [T , L] � M and [F , L] � M; also, if k > j then [Fi, Fk] � Tk � M forall i � k and [Fk, Fl] � Tl � M for all l > k. Thus [L, G] � M and so LG ∩ T = L[L, G] ∩ T � LM ∩ T =M(L ∩ T ). Since each Fi is generated by free generators mod T we have L ∩ T = L′ , which is alsocontained in M , and so LG ∩ T � M and t /∈ MLG . Since G/MLG is finite-by-free abelian there is againa normal subgroup K of finite index in G such that t /∈ K , and so G is residually finite. The proof forhomomorphic images G/N of the type described is essentially identical.

Finally, let H be an arbitrary residually finite group that has a normal nilpotent subgroup J ofclass c such that H/ J is divisible, and let J∗ be a maximal nil-c normal subgroup of H that contains J .Then H/ J∗ is residually finite (by the maximality of J∗) and since it is also divisible it is trivial. Theresult follows. �Lemma 8. If H is an arbitrary subgroup of G that contains F then H has defect at most d in G.

Proof. It suffices to prove that [G, dU ] � U for every finitely generated subgroup U containing F .Let U be such a subgroup and let X/T = IG/T (U T /T ). Since G/T is free abelian we have G/T =X/T × V /T for some subgroup V . By Lemma 7 G is residually finite, and since G is countable thereis a descending chain G1 > G2 > · · · of normal subgroups of finite index in G with trivial intersection.Set V i = V ∩ Gi for each i � 1; thus the V i form a descending chain of G-invariant subgroups of finiteindex in V with trivial intersection. For each i � 1 we have U V i of finite index in U V and hence inG (since T � V and X/U T is finite). Thus U V i /∈ X. Let a ∈ [G, dU ]; then a ∈ ⋂∞

i=1[G, dU V i] ∩ T �⋂∞i=1(U V i) ∩ T .For each i � 1 there are elements ui , vi of U , V i respectively such that a = ui vi . Then, for all i � 1,

u−1i a ∈ V ∩ U T = T , and it follows that u−1

i ∈ U ∩ T , which is finite. Thus there is an infinite set S ofpositive integers such that ui = u j for all i, j ∈ S , and hence an element u of U such that a = uvi foreach i ∈ S . By the choice of the V i we now have u−1a = 1 and hence a ∈ U , as required. �Lemma 9. There is a positive integer r such that, for every element x of G, 〈F , x〉 has rank at most r.

Proof. We imitate for the most part an argument employed in [3]. Let s be the rank of F (whichwe recall is contained in A), let x ∈ G and write X = F 〈x〉 , Y = X p . We see that F Y /Y has orderat most ps , and we may apply Lemma 1 of [3] (to the group X〈x〉/Y ) to deduce that there is aninteger f , depending only on p, s and d, such that [X, f x] � Y ; thus [F , f x] � Y . Now an easy in-duction shows that, for every positive integer n and element a of A, [a, xn] ∈ 〈[a, j x]: j = 1, . . . ,n〉,and for n = f this tells us that [F , 〈x〉] is generated by at most s( f − 1) elements modulo Y . Settingr = sf + 1, we obtain the desired result. �Lemma 10. Let X be an arbitrary subgroup of G. Then G/[T , X] is residually finite, and in the case where X isfinitely generated we have that G/[T , X] /∈ X and if G = 〈X, Y 〉T for some subgroup Y then T Y /∈ X.

Proof. First note that [Ti, X] � G for all i because [T , G ′] = 1. Since [T , X] is the direct productof the subgroups [Ti, X] it follows from Lemma 7 that G/[T , X] is residually finite. If X is finitely

H. Smith / Journal of Algebra 347 (2011) 133–142 141

generated then T X is nilpotent and so G/(T X)′ /∈ X, by Lemma 1, which implies G/[T , X] /∈ X. IfG = 〈X, Y 〉T and T Y ∈ X (again with X finitely generated) then X is nilpotent and subnormal in Gand, by Lemma 7, T Y is nilpotent. Thus G = (T Y )X is nilpotent, a contradiction. �

In view of Lemma 7 we have that every residually finite non-nilpotent section of G is a coun-terexample to the theorem, and this will allow us to dispense with some of the conditions on G andconsider instead a section of G that has an even more transparent structure. Continuing with our in-vestigation of the group G described at the beginning of this section, we note next that every sectionof G that is k-Engel for some integer k is in fact nilpotent [22].

There is certainly an element g1 of G and an element a1 of T such that [a1, g1] �= 1, and wemay choose a1 to lie in Ti1 and g1 to lie in Q 1 := 〈F1, . . . , Fi1 〉 for some i1: here we have used thefact that [Ti, F j] = 1 whenever i < j. Set j1 = i1 and write V 1 = 〈Ti: i > j1〉. We deduce easily fromLemmas 7 and 10 that G/[V 1, Q 1] is residually finite and not in X (equivalently, it is not nilpotent)and, putting R1 = 〈Fi: i > j1〉, we see that R1 V 1/[V 1, Q 1] is not nilpotent. Hence there are elementsg2 of R1,a2 of V 1 such that [a2, 2 g2] /∈ [V 1, Q 1], and we may suppose that a2 ∈ Ti2 and g2 ∈ Q 2 :=〈F j1+1, . . . , Fi2 〉, for some i2 > j1 = i1.

For each l � i2, let Kl = 〈Ti: i > l〉; then the Kl form a descending series of subgroups of fi-nite index in V 1 and (

⋂∞l=i1

)[V 1, Q 1]Kl = [V 1, Q 1] (as V 1/[V 1, Q 1] is just the direct product ofthe finite subgroups Ti[V 1, Q 1]/[V 1, Q 1]), and so we may choose j2 � i2 such that [a2, 2 g2] /∈[V 1, Q 1]V 2, where V 2 = K j2 (= 〈Ti: i > j2〉). Let R2 = 〈Fi: i > j2〉; then R2 V 2 is not nilpotent modulo[V 1, Q 1][V 2, Q 2] and so there are elements g3 of R2,a3 of V 2 such that [a3, 3 g3] /∈ [V 1, Q 1][V 2, Q 2],and we may suppose that a3 ∈ Ti3 and g3 ∈ Q 3 := 〈F j2+1, . . . , Fi3 〉, for some i3 > j2(� i2). Then thereis an integer j3 � i3 such that [a3, 3 g3] /∈ [V 1, Q 1][V 2, Q 2]V 3, where V 3 = 〈Ti: i > j3〉.

We may continue with this process and obtain the following.

Infinite sequences of positive integers i1 < i2 < · · · ; j1 < j2 · · · such that j1 = i1 and jl � il > jl−1 for alll � 2.Finite G-invariant subgroups Til , l = 1,2, . . . , chosen from among the direct factors Ti of T , and a descend-ing chain of subgroups V 1 > V 2 > · · ·, where Vl := 〈Ti: i > jl〉 of T .Finitely generated subgroups Q l of G given by Q 1 = 〈Fi : i = 1, . . . , i1〉 and, for l � 2, Q l = 〈Fi: jl−1 <

i � il〉.For each l � 1, an element al of Til and an element gl of Q l such that [a1, g1] �= 1 and, for l � 2, [al, l gl] /∈〈[V i, Q i]: i = 1, . . . , l − 1〉Vl.

Let W = 〈[V i, Q i]: i � 1〉 and suppose (for a contradiction) that for some l � 2 we have σ :=[al, l gl] ∈ W . Then σ ∈ [V 1, Q 1] . . . [Vl−1, Q l−1]Vl , since [V i, Q i] � Vl for all i � l. This contradicts thechoice of al and gl , and so (for all l � 1), [al, l gl] /∈ W . Now W is a direct product of G-invariantsubgroups of the Ti , i � 1, and so G/W is residually finite, by Lemma 7. Let X be the subgroupgenerated by all Ti such that i �= il for any l � 1, and set M = X W . Then G/M is residually finite andnon-nilpotent since, for every l � 1, we have al ∈ Til � G . Indeed, [al, l gl] /∈ M for any l.

We claim that, for all i, l with i �= il , [Ti, gl] � M . If i �= ik for any k then this is clear, so let usconsider [Tik , gl] for arbitrary k �= l. In the case where l = 1 we have gl = g1 ∈ Q 1 = 〈Fi: 1 � i � i1〉,and since k > 1 in this case we have Tik � V 1 and hence [Tik , g1] � [V 1, Q 1] � M . For l � 2 wehave gl ∈ Q l = 〈Fi: jl−1 < i � il〉. Now Q l centralizes Tik if ik � jl−1 and hence if ik � il−1, that is, ifk � l − 1, and because k �= l we may hence assume that k > l. But ik > jk−1 and Vk−1 = 〈Ti: i > jk−1〉,and now we have Tik � Vk−1 � Vl and so [Tik , gl] � [Vl, Q l] � M , and the claim is established.

We may now factor and hence assume that M = 1. Again denoting by T the torsion subgroup ofour group G , observe that T is just the direct product of the Til , l � 1, and we write Bl = Til for each l.The original condition that [F j, Fk] � Tk whenever j � k now gives us [g j, gk] ∈ Bk for j < k. To seethis, note that, prior to factoring by M := X W , we have (for j < k) [g j, gk] ∈ [Q j, Q k] � 〈Ti: jk−1 <

i � ik〉, while each of these Ti other than Tik is contained in M . Similarly, we have [F , gk] � Bk(mod M) for all k. We now pass to the (residually finite, non-nilpotent) subgroup G∗ of G given byG∗ = (T F )〈gl: l � 1〉.

142 H. Smith / Journal of Algebra 347 (2011) 133–142

Re-labelling this group G we have [Bi, g j] = 1 whenever i �= j and [gi, g j] ∈ B j whenever i < j;also [F , g j] ∈ B j . Recall too from Lemma 9 that 〈F , x〉 has rank at most r for every x ∈ G and fromLemma 8 that every subgroup of G that contains F has defect at most d in G .

For m = 1,2, . . . , let xm = g1 . . . gm . Then [T , dxm] = ∏mi=1[Bi, [d]xm] = ∏m

i=1[Bi, d gi], since[Bi, g j] = 1 when i �= j. Also, [T , dxm] � 〈F , xm〉 and therefore has rank at most r, so we maychoose m so that the rank of [T , dxm] is maximal. For all j > m we have [T , dx j] = ∏m

i=1[Bi, d gi] ×∏ j

i=m+1[Bi, d gi], and it follows that [B j, d g j] = 1 for all j > m. This is a contradiction, and the theo-rem is proved.

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